Question

Prove that \[ 4\left(\cot^{-1} 3 + \cot^{-1} 2 \right) = \pi. \]

Hint:

Use the sum formula for inverse cotangent: \[ \cot^{-1} x + \cot^{-1} y = \cot^{-1} \left(\frac{xy - 1}{x + y}\right) \quad \text{for } xy>1, \] and remember key inverse trigonometric values.

Answer 1

Let: \[ \alpha = \cot^{-1} 3, \quad \beta = \cot^{-1} 2 \] Recall the formula for sum of inverse cotangents: \[ \cot^{-1} x + \cot^{-1} y = \cot^{-1} \left(\frac{xy - 1}{x + y}\right), \quad \text{if } xy>1. \] Calculate: \[ \alpha + \beta = \cot^{-1} 3 + \cot^{-1} 2 = \cot^{-1} \left(\frac{3 \times 2 - 1}{3 + 2}\right) = \cot^{-1} \left(\frac{6 - 1}{5}\right) = \cot^{-1} \left(\frac{5}{5}\right) = \cot^{-1} 1 \] We know: \[ \cot^{-1} 1 = \frac{\pi}{4}. \] Therefore: \[ 4(\cot^{-1} 3 + \cot^{-1} 2) = 4 \times \frac{\pi}{4} = \pi. \]